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A person invest Rs.1000 at $10\%$ annual compound interest for $2$ years$.$ At the end of two years, the whole amount is invested at an annual simple interest of $12\%$ for $5$ years$.$ The total value of the investment finally is $:$

1. $1776$
2. $1760$
3. $1920$
4. $1936$

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Principle $(P) = Rs. 1000,$ Rate$(R) =10\%,$ Time$(T) = 2$ years and compounded annually.

Amount after $1^{st}$ year = $P + \text{Interest} = 1000 + (10\%\text{ of } 1000) = 1000 +100 = Rs.1100.$

Amount after $2^{nd}$ year = $P_{\text{after}\;1^{st}\text{ year}} + \text{Interest} = 1100 + (10\%\text{ of } 1100) = 1100 + 110 = Rs.1210.$

We can also get the final amount using the compound interest formula $P\left(1+\frac{r}{100})^n\right) = 1000 \times (1+0.1)^2 = Rs. 1210$

Now $P= Rs. 1210, R = 12\%, T= 5$ years for SI.

$SI = \left ( \frac{PRT}{100} \right ) = \left ( \frac{1210\times 12\times 5}{100} \right ) = 121\times 6 = 726.$

$Amount = P + SI = 1210 +726 = 1936.$

$\therefore$ Option D is the correct answer.

by Boss (25.1k points)
selected by
+1 vote

Compound interest : $A=P\left ( 1+\dfrac{R}{100} \right )^T$

$\underbrace{|......10...........|........10.........|}$

$2 \ years$

$amount=1000\left ( 1+\dfrac{10}{100} \right )^2=$ $110\times 11$

Simple interest : $A=P\left ( 1+\dfrac{RT}{100} \right )$

$amount=$ $110\times 11$ $\left ( 1+\dfrac{12\times 5}{100} \right )=1936$

by Loyal (5.5k points)